How do you solve $2 log (2 x) = 1 + log a$ $2log(2x) = 1 + loga$ ?

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Answer 1 · 2016-08-15T01:56:03

$x = \frac{\sqrt{10 a}}{2}$ $x=sqrt(10a)/2$ (Assuming $log = {log}_{10}$ $log = log_10$ )

$2 {log}_{10} (2 x) = 1 + {log}_{10} a$ $2log_10(2x) = 1+log_10 a$

$2 {log}_{10} 2 x - {log}_{10} a = 1$ $2log_10 2x - log_10 a =1$

$2 {log}_{10} 2 x - 2 {log}_{10} a^{\frac{1}{2}} = 1$ $2log_10 2x - 2log_10 a^(1/2) =1$

$2 {log}_{10} (\frac{2 x}{\sqrt{a}}) = 1$ $2log_10((2x)/sqrt(a)) =1$

${log}_{10} (\frac{2 x}{\sqrt{a}}) = \frac{1}{2}$ $log_10((2x)/sqrt(a)) =1/2$

$\frac{2 x}{\sqrt{a}} = 10^{\frac{1}{2}}$ $(2x)/sqrt(a) = 10^(1/2)$

$2 x = \sqrt{10 a}$ $2x=sqrt(10a)$

$x = \frac{\sqrt{10 a}}{2}$ $x=sqrt(10a)/2$

How do you solve 2log(2x) = 1 + loga2log(2x)=1+loga2log(2x) = 1 + loga?